Oscillations of fluid in a U tube

[VortexLattice]

Homework Statement

We have a U tube, like this one:

With the a nonviscous, incompressible fluid at height h at equilibrium. We're interested in finding the frequency of small oscillations about the equilibrium. The tubes have area A (though I'm guessing this falls out in the end) and are L apart from each other (this definitely falls out).

Homework Equations

At equilibrium:

f_{app} = \frac{1}{\rho}\nabla p

Where f_app is an applied force per unit mass, \rho is the density (constant here), and p is the pressure.

The Attempt at a Solution

There are two forces at play: The applied gravitational force per unit mass, f_{app} = -g\hat{z}, and the pressure force (I think). At equilibrium, they should be equal. I'm going to say z is in the vertical direction, and z = 0 at equilibrium. Using the equation above and solving for p, we find that p = -\rho g z.

This is good because it's a linear term in z. Now, I know I need to apply Newton's 2nd law and get something of the form \ddot{z} = -k^2 z. We have p, and force on a surface is pA, where A is the area. So I almost have it, but I'm having trouble putting it all together.

I'm also confused because it seems like when I try to sum all the forces for Newton's 2nd law, I have the pressure force and the gravitational force. But the gravitational force doesn't have z in it... I guess I could do that substitution trick where I let x = z - g, but it just seems like I'm doing something simple wrong.

Can anyone help?? Thanks!

 

[technician]

The displaced liquid of height h exerts a pressure =hρg on the rest of the liquid in the U tube.
The force on the liquid is therefore =pressure/area = hρg/A.
This force is proportional to h (displacement... actually 2 x displacement) therefore the system undergoes SHM.
If you work through the logic of SHM you should get an equation for the time period.
I hope this gets you started... ask if you need any more help

 

[RTW69]

Isn't Force= pressure*area?

F=-ρghA where A=area

d²h/dt²=-(ρgA)h/m

but m=ρAh problem?

or are we going to set ω²=-ρgA/m?

d²h/dt²=-ω²h

 

[technician]

SORRY!RTW. My mistake f =P x A.
Acceleration = F/m where m = total mass of liquid in tube
This is the acceleration of the SHM.
Is this any help...sorry again about my careless typing error

 

[RTW69]

OK is freq=√(ρgA/m)/2∏ for this problem?

 

[technician]

That is what I got !

 

[Stonebridge]

OK is freq=√(ρgA/m)/2∏ for this problem?

Not quite.
In the analysis for SHM, the displacement from the equilibrium position is h/2 here
not h. Which means there is a factor of 2 missing in the formula.

Note also that m = ρLA where L is the total length of the liquid in the U-Tube

This gives T = 2∏√(L/2g) and bears a striking resemblance to another well-known formula!
It shows that the period is independent of the density or mass of the liquid.

 

[technician]

I think I now agree with the displacement =h/2.
I think the question is not very clear, I left the mass of the liquid as m in the equation
Cheers